1) Chemical equation
2Al + 6 HCl ---> 2Al Cl3 + 3 H2
2) molar ratios
2 mol Al : 3 moles H2
3) Proportion
2 mol Al / 3mol H2 = x / 9 mol H2
4) Solve for x
x = 9 mol H2 * 2 mol Al / 3 mol H2 = 6 mol Ag
Answer: 6 moles
An aqueous solution contains the following ions Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ and more than one precipitate will form are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.
<h3>What is precipitate?</h3>
Precipitate is the insoluble compound which is present at the bottom of any chemical reaction in the solid state.
If in an aqueous solution Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ ions are present then:
- Compounds AgCl, PbCl₂, PbSO₄ & Ag₂SO₄ are not soluble in water as it is present in the form of precipitate.
- Pb(NO₃)₂ is fully soluble in water and will not make precipitate.
Hence precipitates are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.
To know more about precipitates, visit the below link:
brainly.com/question/2437408
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At 3.5s the distance would be 10.85
Find 1/4th of 2.00 and add that to 6.2
The name come from the German 'Bisemutum' a corruption of 'Weisse Masse' meaning white mass.
[H_{3}O^{+}] = 0.00770 M
The equilibrium equation representing the dissociation of 

Given [H_{3}O^{+}] = 0.00770 M
Let the initial concentration of acid be x and change y
So y =
=
= 0.00770 M



0.00257 x - 0.00001979 = 0.00005929
x = 0.031 M
Therefore, initial concentration of the weak acid is <u>0.031 M</u>